Chapter 25, Problem 26P

To determine

How far, and in what direction (towards or away from the print)did the writer move the lens?

Answer to Problem 26P

Solution:

The distance moved by writer is $3.9\text{cm}$

Explanation of Solution

Given info:

Focal length, $f=12\text{cm}$

Image distance, $v=25\text{cm}$

Object distance

Formula used:

$1/f=1/u+1/v$

Calculation:

As initially the image was at infinity, therefore the object distance $=12\text{cm}$ (coinciding the focus)

For virtual and magnified image, object should be between focus and pole.

So, the writer moves the lens closer to the fine print.

Now, using above equation, by putting the given values to calculate the image distance

$\begin{array}{l}1/f=1/u+1/v\\ 1/12=1/u-1/25\\ 1/u=1/12+1/25\\ u=8.10\text{cm}\end{array}$

Therefore, the distance moved by the writer

$\begin{array}{l}\text{=12}-8.10\\ =3.9\text{cm}\end{array}$